The Laplacian in curvilinear coordinates - the full story

Peter Haggstromwww.gotohaggstrom.com

mathsatbondibeach@gmail.com

July 22, 2018

1 Introduction

In this article I provide some background to Laplace’s equation (and hence the Laplacian ) as well asgiving detailed derivations of the Laplacian in various coordinate systems using several different tech-niques.

So where does Laplace’s equation come from? Following Newton’s work on gravitation and many otherthings the methods of calculus were applied in the 18th and 19th centuries to gravitational, electrostatic,electromagnetic, hydrodynamic, acoustic and many other phenomena. Laplace’s monumental treatise oncelestial mechanics contains the equation ∂2φ

∂x2 +∂2φ∂y2 +

∂2φ∂z2 = 0 in the context of the behaviour of fluids as

the following extract from a translation of the treatise shows ( [6],page 237):

The full derivation of Laplace’s equation in the context of fluids is quite complex the way he did it and

1

can be found at [6, pages 222-237]. A shorter derivation in the context of heat flow can be found in [4,pages 76-78] and runs as follows. Kellogg’s book was used for many years by students at MIT and otheruniversities. I will reproduce the derivation but there is one step the author glosses over which is quitefundamental to an understanding of partial differential equations in a physical modelling context.

It is supposed we have a solid all of whose points are not at the same temperature. The rate of flow ofheat is represented by a vector (u, v,w) whose direction at any point is that in which the heat is flowingand whose magnitude is obtained by taking an element ∆S of the plane through the point P, say, normalto the direction of flow, then determining the number of calories (CGS system) or Joules (SI system ) persecond flowing through the element and dividing this number by the area ∆S and then finally taking thelimit of the resulting quotient as the maximum chord of ∆S approaches 0. In what follows I will stickwith references to calories.

It is usually assumed that the velocity of the heat flow is proportional to the rate of fall of the temper-ature, U, at P. This constant of proportionality depends of the nature of the material and is referred toas conductivity. The next assumption is that the material is thermally isotropic which means that theconductivity has no directional bias. On this basis it is to be expected that the flow vector is in the samedirection as the gradient of the temperature but with the opposite sign (heat flows from warmer places tocooler places). Thus we have:

u = − k∂U∂x

v = − k∂U∂y

w = − k∂U∂z

(1)

where k is the conductivity factor which is assumed constant for homogeneous materials. Note that inLaplace’s derivation of his equation for fluids showcased above, the factor ρ is assumed to be constantfor a homogeneous fluid. The flow field is always normal to the isothermal surfaces U = constant. Thisis because the gradient is normal to such surfaces.

The next physical assumption involves considering a region T in the body and balancing the rate of theflow of the heat into it against the rise in temperature. The rate of flow of heat into T in calories persecond is the negative of the flux of the field (u, v,w) out of the bounding surface i.e:

−

"S

Vn dS = −

"S

(ul + vm + wn) dS (2)

Vn is the component of the velocity of heat flow in the direction of the outward normal to the surface andl,m, n are the respective direction cosines relating the the x, y, z axes. If the specific heat of the material isc then a unit of heat will raise the temperature of a unit mass by c degrees. Hence the number of caloriesper second received per unit mass is:

c∂U∂t

(3)

The number of calories per second received by the whole mass in T is therefore:

2

$T

cρ∂U∂t

dV (4)

where ρ is the density per unit mass of the material.

The Divergence Theorem allows us to assert that:$T

divV dV =

"S

Vn dS (5)

Hence from (2):

"S

Vn dS = −

$T

divV dV

= −

$T

(∂u∂x

+∂v∂y

+∂w∂z

) dV(6)

But: "S

Vn dS =

$T

cρ∂U∂t

dV (7)

Therefore using (6) and (7): $T

(cρ∂U∂t

+∂u∂x

+∂v∂y

+∂w∂z

)dV = 0 (8)

Assuming the integrand of (8) is continuous, the integrand must vanish since the integral vanishes forevery region T . That this is the case follows from the fundamental lemma of the calculus of variationswhich, at it its simplest, says that if f (x) is continuous on [a, b] and if

∫ ba f (x) h(x) dx = for every contin-

uous function h(x) on [a, b] such that h(a) = h(b) = 0, then f (x) = 0 for all x ∈ [a, b]. This logic extendsto three dimensions as is the case in (8). A proof of the one dimensional case can be found in [3, page9]. The gist of the proof that the integrand of (8) vanishes runs as follows. For the purposes of a contra-diction we suppose our region is a small cube in which the integrand is non-zero, say, positive at somepoint. Because of continuity there is a neighbourhood of this point where the integrand is positive andso the integral over this small, arbitrary cube must be positive, thereby producing a contradiction. This isessentially what Hilbert and Courant do in their proof of the one dimensional case [2, page 185].

Thus we have come to this physical assumption:

∂U∂t

= −1cρ

(∂u∂x

+∂v∂y

+∂w∂z

)(9)

Kellogg says the following [ 4,page 78]:

”The flow of heat in a body may be stationary ie such that the temperature at each point is independentof the time. Such, for instance, might be the situation in a bar, wrapped with insulating material, one end

3

of which is kept in boiling water, and the other end in ice-water. Though the heat would be constantlyflowing, the temperature might not vary sensibly with the time”

Let’s just take this for granted at the moment but it deserves much more explanation than this throwawayline.

Substituting (1) into (9) we have (assuming c, k, ρ are constants):

∂U∂t

= −1cρ

(∂

∂x

(− k

∂U∂x

)+∂v∂y

(− k

∂U∂y

)+∂

∂z

(− k

∂U∂z

))=a2

(∂2U∂x2 +

∂2U∂y2 +

∂2U∂z2

) (10)

where a2 = kcρ .

On the assumption of stationarity we have ∂U∂t = 0 and Laplace’s equation results:

∆U =∂2U∂x2 +

∂2U∂y2 +

∂2U∂z2 = 0 (11)

Let us now return to Kellogg’s quoted statement above concerning stationarity ie lack of variation in time.On the surface of things this is a bold claim since one would expect the heat to move over time givingrise to a variation in temperature with respect to time.

The heat equation is such an important equation in physics it is worth understanding some dimensionsto it which are not to my knowledge covered generally in undergraduate engineering courses or indeedphysics courses. What follows is an expansion of some comments made by the well known partial differ-ential equation expert Luis Cafarelli of the University of Texas, Austin in his plenary lecture at the 2013Mathematical Congress of the Americas (MCA) at Guanajuato, Mexico. The one dimensional heat equa-tion ie ∂u

∂t = ∂2u∂x2 represents a diffusion process. A diffusion process such as that represented by the heat

equation has a tendency to revert to its surrounding average. To see how this might be the case we needto look at the most simple situation – ie one dimension, which indicates a relationship between diffusionand the Laplacian (in n dimensions the Laplacian is ∆u =

∑ni=1

∂2u∂x2

i).

In one dimension the Laplacian of u is simply the second derivative of u and so we look at the limit of thesecond order incremental quotient. Recall that:

u′′

(x) = limh→0

u(x+h)−u(x)h −

(u(x)−u(x−h)

h

)h

= limh→0

u(x + h) − 2u(x) + u(x − h)h2 (12)

In the diagram below the balls represent particles which can jump left and right in proportion to theirnumber in a pile. The pile at x gains half of the particles coming from adjacent piles and loses its own.This simple rule gives rise to a balance equation of gains (ie 1

2 u(x − h) + 12 u(x + h)) minus losses (ie u(x)

)which is proportional to:

12

(u(x + h) + u(x − h) − 2u(x)

)(13)

4

Equation (13) looks suspiciously like (12) - hence the connection with the Laplacian which has remark-able features: it is rotationally invariant, independent of the system of coordinates and represents a dif-fusion. As we go up in dimensions we consider the Laplacian as a limit gain-loss of density u at x. Wetake the average over a unit sphere S of the radial second derivatives in every direction and one of thefundamental results of harmonic analysis is that:

∆(u) =

?S

urr dA(s) (14)

Recall that for a function u defined in a ball B(x, r) of radius r about x in Rn, with boundary ∂B(x, r) andα(n) is the volume of a unit ball in Rn and nα(n) is the surface area of the unit ball in Rn, the average of uon B(x, r) is: ?

B(x,r)u(y) dy =

1α(n)rn

∫B(x,r)

u(y) dy (15)

In 2 dimensions a function u is harmonic at P (ie it satisfies Laplace’s equation ∆ = 0) if and onlyif:

u(P) =1

2πr

∫∂B(P,r)

u ds =1πr2

∫B(P,r)

u dx dy (16)

To prove (16) we take P = (x0, y0) and we suppose that u(x0, y0) = 12πr

∫∂B(P,r) u(x, y) ds then:

u(x0, y0) =1

2πr

∫∂B(P,r)

u(x, y) ds =1

2πr

∫ 2π

0u(x0 + r cos θ, y0 + r sin θ) rdθ

=1

2π

∫ 2π

0u(x0 + r cos θ, y0 + r sin θ) dθ (17)

The LHS of (17) is simply a constant so if we differentiate with respect to r under the integral sign (anduse the chain rule) we get:

0 =1

2π

∫ 2π

0(∂u∂x

cos θ +∂u∂y

sin θ) dθ =1

2πr

∫ 2π

0(∂u∂x

cos θ +∂u∂y

sin θ) rdθ

=1

2πr

∫∂B(P,r)

∇u ν ds =1

2πr

∫B(P,r)

div (∇u) dy dx =1

2πr

∫B(P,r)

∆u dy dx (18)

5

The divergence theorem justifies the last step in (18). Hence based on our assumption u(x0, y0) =1

2πr

∫∂B(P,r) u(x, y) ds we have shown that 0 =

∫B(P,r) ∆u dy dx for all r > 0. If all this holds for every P

in some open subset Ω in R2 then we must have that ∆u = 0 for each such P. Thus u is harmonic.

What this averaging suggests is that the heat equation ∂u∂t = ∆(u) reflects the fact that the density u

at the point x makes an infinitesimal comparison within its neighbourhood and tries to revert to thesurrounding average.

It is this local behaviour that justifies Kellogg’s claim concerning stationarity which, at first blush, seemscounterintuitive.

2 The taxonomy of PDEs

In the taxonomy of partial differential equations (PDEs) the broad breakdown is:

Laplace’s equation: ∆u = ∂2u∂x2 + ∂2u

∂y2 = 0 Elliptic

Wave equation: ∂2u∂t2 − c2 ∂2u

∂x2 = 0 Hyperbolic

Heat equation: : ∂u∂t − γ

∂2u∂x2 = 0 Parabolic

Hyperbolic and parabolic PDEs typically model dynamic phenomena and hence have a time dimension.PDEs which model equilibrium phenomena (where time has in effect decayed away) are generally ellipticin nature and involve only spatial dimensions as in the case of the Laplace and Poisson equations. EllipticPDEs are associated with boundary value problems whereas parabolic and hyperbolic PDEs are associatedwith initial value and initial boundary value problems ([7] section 4.4)

The classification of linear second order PDEs arises from the form of the “discriminant” of the generalform of the PDE. Writing uxx for ∂2u

∂x2 etc the general form of a linear second order PDE in two variablesx and y is:

L[u] = Auxx + Buxy + Cuyy + Dux + Euy + Fu = G (19)

A,B,. . . G can all be functions of x and y and if G ≡ 0 the PDE is homogeneous.

What drives the classification is the discriminant which is defined as:

∆ = B2–4AC (20)

The discriminant is connected with the general quadratic equation where it is assumed that not all of thecoefficients are zero:

Q(x, y) = Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (21)

The classification becomes:

Hyperbolic if ∆ > 0

Parabolic if ∆ = 0

6

Elliptic if ∆ < 0

Equation (21) can be written in matrix form as:

Xt MX + NX + F = 0 (22)

where the matrix M has the form:

M =

(A B

2B2 C

)(23)

By the Spectral Theorem (see [1] pages 256-257) there is an orthogonal matrix P such that PMPt diag-onalises M and one can then apply a translation to eliminate as much as possible the linear and constantterms NX + F. A change of variable X′ = PX or X = PtX

′

is then made. The equation is thus reduced toone of the canonical forms for an ellipse, hyperbola or parabola. Note that −4 det M = B2 − 4AC.

Rather than using x, y etc we can formulate the general linear PDE as follows:

n∑i, j=1

ai juxi x j︸ ︷︷ ︸Principal Part

+

n∑i, j=1

biuxi + cu = d (24)

Assuming continuity of the second partials we will have uxi x j = ux j xi and the n × n matrix A = (ai j) willbe symmetric and so ai j = a ji. We can then diagonalise the principal part of (24).

As an example we can find an orthogonal transformation which eliminates the mixed partials in thefollowing equation:

2ux1 x1 + 2ux2 x2 − 15ux3 x3 + 8ux1 x2 − 12ux2 x3 − 12ux1 x3 = 0 (25)

The matrix corresponding to the principal part of (25) is:

A =

2 4 −64 2 −6−6 −6 −15

(26)

The eigenvalues of A are −2,−18 and 9 and the orthogonal matrix B we want is found to be:

B =

1√

21

3√

223

−1√

21

3√

223

0 43√

2−13

(27)

Thus we will have:

C = BtAB =

−2 0 00 −18 00 0 9

(28)

7

If we make a change of variables ξ = Btx (25) transforms to:

− 2uξ1ξ1 − 18uξ2ξ2 + 9uξ3ξ3 = 0 (29)

We then let:

η1 =ξ1√

2

η2 =ξ2

3√

2

η3 =ξ3

3

(30)

With these changes and multiplying (29) by −1 and noting that uξiξi = ∂∂ξi

(∂u∂ξi

)etc we get:

uη1η1 + uη2η2 − uη3η3 = 0 (31)

This is an hyperbolic PDE. More details are set out in the appendix.

3 The Laplacian in polar coordinates – brute force

There are various ways of deriving the Laplacian in curvilinear coordinates such a polar, cylindrical orspherical. This is a purely mathematical exercise unrelated to the underlying physics of, say, electostatics.The first method is simply one of brute force and it gets increasingly more complicated as we movefrom polar, to cylindrical and then spherical coordinates. There are many other curvilinear coordinatesystems beyond those just mentioned eg parabolic cylindrical, paraboloidal, elliptic cylindrical etc. Thereis nothing deep about the process but it is error prone. It also suffers from a lack of intuition which makesit hard to remember the form of the Laplacian. Later on I show how to remember a general formula forthe Laplacian.

A second method is favoured in a physics and engineering contexts and it involves divergences anddirectional derivatives. This approach is followed in [8] and many other textbooks.

The third method is completely general and has it roots in tensor calculus/differential geometry.

In this article I systematically go through each method.

In applying the chain rule it is crucial to remember that the partial derivatives are calculated holding othervariables constant. Because the chain rule is generally written without explicit reference to the variablesbeing held constant it is very easy to make errors that will build upon one another. Thus, for instance, thechain rule can be written explicitly this way:

(∂u∂x

)y constant

=(∂u∂r

)θ constant

( ∂r∂x

)y constant

+(∂u∂θ

)r constant

(∂θ∂x

)y constant

(32)

∂r∂x and ∂θ

∂x are evaluated at constant y and ∂r∂y and ∂θ

∂y are evaluated at constant x.

8

The trap is to do this.

x =r cos θ

=⇒ r =x

cos θ

=⇒∂r∂x

=1

cos θ

(33)

This is actually(∂r∂x

)θ constant

. Keep these principles in mind in what follows.

The relevant equations are:

x = r cos θy = r sin θ

(34)

θ = arctan( y

x

)r =

√x2 + y2

(35)

In preparation for applying the chain rule we now need to derive the following derivatives and we assumeequality of mixed partials:

∂r∂x

=12

(x2 + y2)−12 × 2x =

xr

= cos θ (36)

∂r∂y

=yr

= sin θ (37)

∂θ

∂x=

1

1 +y2

x2

×−yx2 =

−yr2 =

− sin θr

(38)

9

∂θ

∂y=

1

1 +y2

x2

×1x

=xr2 =

cos θr

(39)

We now need to calculate ∂u∂x and then ∂2u

∂x2 and similarly for ∂u∂y and then ∂2u

∂y2 .

∂u∂x

=∂u∂r

∂r∂x

+∂u∂θ

∂θ

∂x

= cos θ∂u∂r−

sin θr

∂u∂θ

(40)

Hence:

∂u2

∂x2 =∂

∂x

(∂u∂x

)=∂

∂r

(∂u∂x

) ∂r∂x

+∂

∂θ

(∂u∂x

)∂θ∂x

= cos θ∂

∂r

[cos θ

∂u∂r−

sin θr

∂u∂θ

]−

sin θr

∂

∂θ

[cos θ

∂u∂r−

sin θr

∂u∂θ

]= cos θ

[cos θ

∂2u∂r2 −

sin θr

∂2u∂r∂θ

+sin θr2

∂u∂θ

]−

sin θr

[− sin θ

∂u∂r

+ cos θ∂2u∂r∂θ

−sin θ

r∂2u∂θ2 −

cos θr

∂u∂θ

]= cos2 θ

∂2u∂r2 −

2 sin θ cos θr

∂2u∂r∂θ

+2 sin θ cos θ

r2

∂u∂θ

+sin2 θ

r∂u∂r

+sin2 θ

r∂2u∂θ2

(41)

We now replicate the process for ∂2u∂y2 .

∂u∂y

=∂u∂r∂r∂y

+∂u∂θ

∂θ

∂y

= sin θ∂u∂r

+cos θ

r∂u∂θ

(42)

10

∂2u∂y2 =

∂

∂y

(∂u∂y

)=

∂

∂r

(∂u∂y

)∂r∂y

+∂

∂θ

(∂u∂y

)∂θ∂y

= sin θ∂

∂r

(∂u∂y

)+

cos θr

∂

∂θ

(∂u∂y

)= sin θ

∂

∂r

(sin θ

∂u∂r

+cos θ

r∂u∂θ

)+

cos θr

∂

∂θ

(sin θ

∂u∂r

+cos θ

r∂u∂θ

)= sin2 θ

∂2u∂r2 +

sin θ cos θr

∂2u∂r∂θ

−sin θ cos θ

r2

∂u∂θ

+sin θ cos θ

r∂2u∂r∂θ

+cos2 θ

r∂u∂r

+cos2 θ

r2

∂2u∂θ2

−sin θ cos θ

r2

∂u∂θ

= sin2 θ∂2u∂r2 +

2 sin θ cos θr

∂2u∂r∂θ

−2 sin θ cos θ

r2

∂u∂θ

+cos2 θ

r∂u∂r

+cos2 θ

r2

∂2u∂θ2

(43)

So adding (41) and (43) we have:

∂2u∂x2 +

∂2u∂y2 = cos2 θ

∂2u∂r2 −

2 sin θ cos θr

∂2u∂r∂θ

+2 sin θ cos θ

r2

∂u∂θ

+sin2 θ

r∂u∂r

+sin2 θ

r2

∂2u∂θ2

+ sin2 θ∂2u∂r2 +

2 sin θ cos θr

∂2u∂r∂θ

−2 sin θ cos θ

r2

∂u∂θ

+cos2 θ

r∂u∂r

+cos2 θ

r2

∂2u∂θ2

=∂2u∂r2 +

1r∂u∂r

+1r2

∂2u∂θ2

(44)

So the Laplacian operator in polar form can be written as follows:

∆polar =∂2

∂r2 +1r∂

∂r+

1r2

∂2

∂θ2 (45)

4 The Laplacian in spherical coordinates - brute force

There is nothing in principle difficult about this but it is incredibly easy to make a slip. You will only dothis once in your life and you will appreciate some of the other methods covered later in this article. Westart with the basic coordinate relationships and list the relevant derivatives:

x = r sin θ cos φy = r sin θ sin φ

z = r cos θ(46)

11

r =

√x2 + y2 + z2

θ = arccos(zr

)

φ = arctan(yx

)

(47)

The Laplacian in rectangular coordinates is:

∆u =∂2u∂x2 +

∂2u∂y2 +

∂2u∂z2 (48)

You will also see ∇2 used as the symbol for the Laplacian.

We now get each component of (48):

∂u∂x

=∂u∂r

∂r∂x

+∂u∂θ

∂θ

∂x+∂u∂φ

∂φ

∂x(49)

∂u∂y

=∂u∂r∂r∂y

+∂u∂θ

∂θ

∂y+∂u∂φ

∂φ

∂y(50)

∂u∂z

=∂u∂r∂r∂z

+∂u∂θ

∂θ

∂z+∂u∂φ

∂φ

∂z(51)

We now need the various sub-components such as ∂r∂x ,

∂r∂y etc and then we will need to work out the second

derivatives. The basic components are given below:

∂r∂x

=2x

2√

x2 + y2 + z2=

xr

= sin θ cos φ (52)

∂r∂y

=yr

= sin θ sin φ (53)

∂r∂z

= cos θ (54)

12

∂θ

∂x=

∂

∂xarccos(

zr

) =∂

∂xarccos(

z√x2 + y2 + z2

)

=−1√

1 − z2

x2+y2+z2

×−2zx

2(x2 + y2 + z2)32

=zx

r2√

x2 + y2

=r2 sin θ cos θ cos φ

r2√

r2 sin2 θ cos2 φ + r2 sin2 θ sin2 φ

=1r

cos θ cos φ

(55)

∂θ

∂y=

∂

∂yarccos(

zr

) =∂

∂yarccos(

z√x2 + y2 + z2

)

=−1√

1 − z2

x2+y2+z2

×−2zy

2(x2 + y2 + z2)32

=zy

r2√

x2 + y2

=r2 sin θ cos θ sin φ

r2√

r2 sin2 θ cos2 φ + r2 sin2 θ sin2 φ

=1r

cos θ sin φ

(56)

∂θ

∂z=∂

∂zarccos(

zr

) =∂

∂zarccos

(z√

x2 + y2 + z2

)=

−1√1 − z2

x2+y2+z2

×x2 + y2

r3

=−(x2 + y2)

r2√

x2 + y2

=−1r

sin θ

(57)

∂φ

∂x=

∂

∂xarctan(

yx

)

=1

1 +y2

x2

×−yx2

=−y

x2 + y2

=− sin φr sin θ

(58)

13

∂φ

∂y=

∂

∂yarctan(

yx

)

=1

1 +y2

x2

×1x

=x

x2 + y2

=cos φr sin θ

(59)

∂φ

∂z= 0 (60)

We start with ∂u∂x in terms of the derivatives with respect to the spherical coordinates (using (52)-(60)

):

∂u∂x

=∂u∂r

∂r∂x

+∂u∂θ

∂θ

∂x+∂u∂φ

∂φ

∂x

= sin θ cos φ∂u∂r

+1r

cos θ cos φ∂u∂θ−

1r

sin φsin θ

∂u∂φ

(61)

Now we work out ∂2u∂x2 :

∂2u∂x2 =

∂

∂r

(∂u∂x

)∂r∂x

+∂

∂θ

(∂u∂x

)∂θ

∂x+

∂

∂φ

(∂u∂x

)∂φ

∂x

= sin θ cos φ∂

∂r

(∂u∂x

)︸ ︷︷ ︸

A

+1r

cos θ cos φ∂

∂θ

(∂u∂x

)︸ ︷︷ ︸

B

−1r

sin φsin θ

∂

∂φ

(∂u∂x

)︸ ︷︷ ︸

C

(62)

A = sin θ cos φ∂

∂r

[sin θ cos φ

∂u∂r

+1r

cos θ cos φ∂u∂θ−

1r

sin φsin θ

∂u∂φ

]= sin θ cos φ

[sin θ cos φ

∂2u∂r2 +

1r

cos θ cos φ∂2u∂r∂θ

−1r2 cos θ cos φ

∂u∂θ

+1r2

sin φsin θ

∂u∂φ−

1r

sin φsin θ

∂2u∂r∂φ

]= sin2 θ cos2 φ

∂2u∂r2 +

1r

sin θ cos θ cos2 φ∂2u∂r∂θ

−1r2 sin θ cos θ cos2 φ

∂u∂θ

+1r2 sin φ cos φ

∂u∂φ

−1r

sin φ cos φ∂2u∂r∂φ

(63)

Hence:

A = sin2 θ cos2 φ∂2u∂r2 +

1r

sin θ cos θ cos2 φ∂2u∂r∂θ

−1r2 sin θ cos θ cos2 φ

∂u∂θ

+1r2 sin φ cos φ

∂u∂φ

−1r

sin φ cos φ∂2u∂r∂φ

14

B =1r

cos θ cos φ∂

∂θ

(∂u∂x

)=

1r

cos θ cos φ∂

∂θ

[sin θ cos φ

∂u∂r

+1r

cos θ cos φ∂u∂θ−

1r

sin φsin θ

∂u∂φ

]=

1r

cos θ cos φ[

cos θ cos φ∂u∂r

+ sin θ cos φ∂2u∂θ∂r

−1r

sin θ cos φ∂u∂θ

+1r

cos θ cos φ∂2u∂θ2

−1r

sin φsin θ

∂2u∂θ∂φ

+1r

sin φ cos θsin2 θ

∂u∂φ

]=

1r

cos2 θ cos2 φ∂u∂r

+1r

sin θ cos θ cos2 φ∂2u∂θ∂r

−1r2 sin θ cos θ cos2 φ

∂u∂θ

+1r2 cos2 θ cos2 φ

∂2u∂θ2

−1r2

cos θ sin φ cos φsin θ

∂2u∂θ∂φ

+1r2

cos2 θ sin φ cos φsin2 θ

∂u∂φ

(64)

Hence:

B =1r

cos2 θ cos2 φ∂u∂r

+1r

sin θ cos θ cos2 φ∂2u∂θ∂r

−1r2 sin θ cos θ cos2 φ

∂u∂θ

+1r2 cos2 θ cos2 φ

∂2u∂θ2

−1r2

cos θ sin φ cos φsin θ

∂2u∂θ∂φ

+1r2

cos2 θ sin φ cos φsin2 θ

∂u∂φ

C = −1r

sin φsin θ

∂

∂φ

(∂u∂x

)= −

1r

sin φsin θ

∂

∂φ

[sin θ cos φ

∂u∂r

+1r

cos θ cos φ∂u∂θ−

1r

sin φsin θ

∂u∂φ

]= −

1r

sin φsin θ

[− sin θ sin φ

∂u∂r

+ sin θ cos φ∂2u∂φ∂r

−1r

cos θ sin φ∂u∂θ

+1r

cos θ cos φ∂2u∂φ∂θ

−1r

cos φsin θ

∂u∂φ−

1r

sin φsin θ

∂2u∂φ2

]=

1r

sin2 φ∂u∂r−

1r

sin φ cos φ∂2u∂φ∂r

+sin2 φ cos θ

r2 sin θ∂u∂θ−

sin φ cos φ cos θr2 sin θ

∂2u∂φ∂θ

+sin φ cos φr2 sin2 θ

∂u∂φ

+sin2 φ

r2 sin2 θ

∂2u∂φ2

(65)

Hence:

C =1r

sin2 φ∂u∂r−

1r

sin φ cos φ∂2u∂φ∂r

+sin2 φ cos θ

r2 sin θ∂u∂θ−

sin φ cos φ cos θr2 sin θ

∂2u∂φ∂θ

+sin φ cos φr2 sin2 θ

∂u∂φ

+sin2 φ

r2 sin2 θ

∂2u∂φ2

We now repeat the process to arrive at ∂2u∂y2 .

15

∂u∂y

=∂u∂r∂r∂y

+∂u∂θ

∂θ

∂y+∂u∂φ

∂φ

∂y

= sin θ sin φ∂u∂r

+1r

cos θ sin φ∂u∂θ

+cos φr sin θ

∂u∂φ

(66)

As before we have:

∂2u∂y2 =

∂

∂r

(∂u∂y

)∂r∂y

+∂

∂θ

(∂u∂y

)∂θ∂y

+∂

∂φ

(∂u∂y

)∂φ∂y

= sin θ sin φ∂

∂r

(∂u∂y

)︸ ︷︷ ︸

D

+1r

cos θ sin φ∂

∂θ

(∂u∂y

)︸ ︷︷ ︸

E

+cos φr sin θ

∂

∂φ

(∂u∂y

)︸ ︷︷ ︸

F

(67)

Now we work out each component:

D = sin θ sin φ∂

∂r

[sin θ sin φ

∂u∂r

+1r

cos θ sin φ∂u∂θ

+cos φr sin θ

∂u∂φ

]= sin θ sin φ

[sin θ sin φ

∂2u∂r2 −

1r2 cos θ sin φ

∂u∂θ

+1r

cos θ sin φ∂2u∂r∂θ

−cos φ

r2 sin θ∂u∂φ

+cos φr sin θ

∂2u∂r∂φ

]= sin2 θ sin2 φ

∂2u∂r2 −

1r2 sin θ cos θ sin2 φ

∂u∂θ

+1r

sin θ cos θ sin2 φ∂2u∂r∂θ

−1r2 sin φ cos φ

∂u∂φ

+1r

sin φ cos φ∂2u∂r∂φ

(68)

So:

D = sin2 θ sin2 φ∂2u∂r2 −

1r2 sin θ cos θ sin2 φ

∂u∂θ

+1r

sin θ cos θ sin2 φ∂2u∂r∂θ

−1r2 sin φ cos φ

∂u∂φ

+1r

sin φ cos φ∂2u∂r∂φ

16

E =1r

cos θ sin φ∂

∂θ

(∂u∂y

)=

1r

cos θ sin φ∂

∂θ

[sin θ sin φ

∂u∂r

+1r

cos θ sin φ∂u∂θ

+cos φr sin θ

∂u∂φ

]=

1r

cos θ sin φ[

cos θ sin φ∂u∂r

+ sin θ sin φ∂2u∂θ∂r

−1r

sin θ sin φ∂u∂θ

+1r

cos θ sin φ∂2u∂θ2 +

cos φr sin θ

∂2u∂θ∂φ

−cos φ cos θ

r sin2 θ

∂u∂φ

]=

1r

cos2 θ sin2 φ∂u∂r

+1r

cos θ sin θ sin2 φ∂2u∂θ∂r

−1r2 cos θ sin θ sin2 φ

∂u∂θ

+1r2 cos2 θ sin2 φ

∂2u∂θ2

+cos θ sin φ cos φ

r2 sin θ∂2u∂θ∂φ

−cos2 θ sin φ cos φ

r2 sin2 θ

∂u∂φ

(69)

Hence:

E =1r

cos2 θ sin2 φ∂u∂r

+1r

cos θ sin θ sin2 φ∂2u∂θ∂r

−1r2 cos θ sin θ sin2 φ

∂u∂θ

+1r2 cos2 θ sin2 φ

∂2u∂θ2

+cos θ sin φ cos φ

r2 sin θ∂2u∂θ∂φ

−cos2 θ sin φ cos φ

r2 sin2 θ

∂u∂φ

F =cos φr sin θ

∂

∂φ

(∂u∂y

)=

cos φr sin θ

∂

∂φ

[sin θ sin φ

∂u∂r

+1r

cos θ sin φ∂u∂θ

+cos φr sin θ

∂u∂φ

]=

cos φr sin θ

[sin θ cos φ

∂u∂r

+ sin θ sin φ∂2u∂φ∂r

+1r

cos θ cos φ∂u∂θ

+1r

cos θ sin φ∂2u∂φ∂θ

−sin φr sin θ

∂u∂φ

+cos φr sin θ

∂2u∂φ2

]=

1r

cos2 φ∂u∂r

+1r

sin φ cos φ∂2u∂φ∂r

+cos θ cos2 φ

r2 sin θ∂u∂θ

+cos θ sin φ cos φ

r2 sin θ∂2u∂φ∂θ

−sin φ cos φr2 sin2 θ

∂u∂φ

+cos2 φ

r2 sin2 θ

∂2u∂φ2

(70)

So:F =

1r

cos2 φ∂u∂r

+1r

sin φ cos φ∂2u∂φ∂r

+cos θ cos2 φ

r2 sin θ∂u∂θ

+cos θ sin φ cos φ

r2 sin θ∂2u∂φ∂θ

−sin φ cos φr2 sin2 θ

∂u∂φ

+cos2 φ

r2 sin2 θ

∂2u∂φ2

Finally we calculate ∂2u∂z2 :

17

∂u∂z

=∂u∂r∂r∂z

+∂u∂θ

∂θ

∂z+∂u∂φ

∂φ

∂z︸︷︷︸=0

=∂u∂r∂r∂z

+∂u∂θ

∂θ

∂z

= cos θ∂u∂r−

1r

sin θ∂u∂θ

(71)

Hence:

∂2u∂z2 =

∂

∂r

(∂u∂z

)∂r∂z

+∂

∂θ

(∂u∂z

)∂θ∂z

= cos θ∂

∂r

(∂u∂z

)︸ ︷︷ ︸

G

−1r

sin θ∂

∂θ

(∂u∂z

)︸ ︷︷ ︸

H

(72)

G = cos θ∂

∂r

(∂u∂z

)= cos θ

∂

∂r

[cos θ

∂u∂r−

1r

sin θ∂u∂θ

]= cos θ

[cos θ

∂2u∂r2 +

1r2 sin θ

∂u∂θ−

1r

sin θ∂2u∂r∂θ

]= cos2 θ

∂2u∂r2 +

1r2 sin θ cos θ

∂u∂θ−

1r

sin θ cos θ∂2u∂r∂θ

(73)

So:

G = cos2 θ∂2u∂r2 +

1r2 sin θ cos θ

∂u∂θ−

1r

sin θ cos θ∂2u∂r∂θ

H = −1r

sin θ∂

∂θ

(∂u∂z

)= −

1r

sin θ∂

∂θ

[cos θ

∂u∂r−

1r

sin θ∂u∂θ

]= −

1r

sin θ[− sin θ

∂u∂r

+ cos θ∂2u∂θ∂r

−1r

cos θ∂u∂θ−

1r

sin θ∂2u∂θ2

]=

1r

sin2 θ∂u∂r−

1r

sin θ cos θ∂2u∂θ∂r

+1r2 sin θ cos θ

∂u∂θ

+1r2 sin2 θ

∂2u∂θ2

(74)

H =1r

sin2 θ∂u∂r−

1r

sin θ cos θ∂2u∂θ∂r

+1r2 sin θ cos θ

∂u∂θ

+1r2 sin2 θ

∂2u∂θ2

Thus ∆spherical = A + B + C + D + E + F + G + H and to see the final result of this we collect coefficientsof the various derivatives (and assume equality of mixed partials):

18

Coefficient of∂u∂θ

=−2r2 sin θ cos θ cos2 φ︸ ︷︷ ︸

A and B

+sin2 φ cos θ

r2 sin θ︸ ︷︷ ︸C

−1r2 sin θ cos θ sin2 φ︸ ︷︷ ︸

D

−1r2 cos θ sin θ sin2 φ︸ ︷︷ ︸

E

+cos θ cos2 φ

r2 sin θ︸ ︷︷ ︸F

+2r2 sin θ cos θ︸ ︷︷ ︸

G and H

=−2 sin2 θ cos θ cos2 φ + sin2 φ cos θ − sin2 θ cos θ sin2 φ − cos θ sin2 θ sin2 φ

r2 sin θ

+cos θ cos2 φ + 2 sin2 θ cos θ

r2 sin θ

=−2 sin2 θ cos θ cos2 φ + cos θ − sin2 θ cos θ sin2 φ − cos θ sin2 θ sin2 φ

r2 sin θ

+2 sin2 θ cos θ

r2 sin θ

=2 sin2 θ cos θ sin2 φ + cos θ − sin2 θ cos θ sin2 φ − cos θ sin2 θ sin2 φ

r2 sin θ

=sin2 θ cos θ sin2 φ + cos θ − sin2 θ cos θ sin2 φ

r2 sin θ

=cot θ

r2

(75)

Coefficient of∂u∂φ

=1r2 sin φ cos φ︸ ︷︷ ︸

A

+cos2 θ sin φ cos φ

r2 sin2 θ︸ ︷︷ ︸B

+sin φ cos φr2 sin2 θ︸ ︷︷ ︸

C

−1r2 sin φ cos φ︸ ︷︷ ︸

D

−cos2 θ sin φ cos φ

r2 sin2 θ︸ ︷︷ ︸E

−sin φ cos φr2 sin2 θ︸ ︷︷ ︸

F

= 0

(76)

Coefficient of∂2u∂r∂θ

=sin θ cos θ cos2 φ

r︸ ︷︷ ︸A

+sin θ cos θ cos2 φ

r︸ ︷︷ ︸B

+sin θ cos θ sin2 φ

r︸ ︷︷ ︸D

+cos θ sin θ sin2 φ

r︸ ︷︷ ︸E

−2 sin θ cos θ

r︸ ︷︷ ︸G and H

=2 sin θ cos θ cos2 φ + 2 sin θ cos θ sin2 φ − 2 sin θ cos θ

r= 0

(77)

Coefficient of∂2u∂r∂φ

=

︷ ︸︸ ︷sin φ cos φ

D

︷ ︸︸ ︷− sin φ cos φ

A

︷ ︸︸ ︷+ sin φ cos φ

F

︷ ︸︸ ︷− sin φ cos φ

C

r= 0

(78)

19

Coefficient of∂2u∂r2 = sin2 θ cos2 φ︸ ︷︷ ︸

A

+ sin2 θ sin2 φ︸ ︷︷ ︸D

+ cos2 θ︸ ︷︷ ︸G

= 1

(79)

Coefficient of∂2u∂θ2 =

1r2 cos2 θ cos2 φ︸ ︷︷ ︸

B

+1r2 cos2 θ sin2 φ︸ ︷︷ ︸

E

+1r2 sin2 θ︸ ︷︷ ︸

H

=1r2

(80)

Coefficient of∂2u∂φ2 =

sin2 φ

r2 sin2 θ︸ ︷︷ ︸C

+cos2 φ

r2 sin2 θ︸ ︷︷ ︸F

=1

r2 sin2 θ

(81)

Coefficient of∂u∂r

=cos2 θ cos2 φ

r︸ ︷︷ ︸B

+sin2 φ

r︸ ︷︷ ︸C

+cos2 θ sin2 φ

r︸ ︷︷ ︸E

= +cos2 φ

r︸ ︷︷ ︸F

+sin2 θ

r︸ ︷︷ ︸H

=1r

+cos2 θ

r(cos2 φ + sin2 φ) +

sin2 θ

r

=2r

(82)

Coefficient of∂2u∂θ∂φ

=− cos θ sin φ cos φ

r2 sin θ︸ ︷︷ ︸B

− sin φ cos φ cos θr2 sin θ︸ ︷︷ ︸

C

+cos θ sin φ cos φ

r2 sin θ︸ ︷︷ ︸E

= +cos θ sin φ cos φ

r2 sin θ︸ ︷︷ ︸F

= 0

(83)

Putting this all together we finally get what we were after:

∆ spherical u =∂2u∂r2 +

2r∂u∂r

+1

r2 sin2 θ

∂2u∂φ2 +

1r2

∂2u∂θ2 +

cot θr2

∂u∂θ

(84)

20

There are equivalent forms of (84) which are as follows:

∆ spherical u =1r2

∂

∂r

(r2 ∂u∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂φ2 (85)

This is verified as follows:

∆ spherical u =1r2

∂

∂r

(r2 ∂u∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂φ2

=1r2

[r2 ∂

2u∂r2 + 2r

∂u∂r

]+

1r2 sin θ

[sin θ

∂2u∂θ2 + cos θ

∂u∂θ

]+

1r2 sin2 θ

∂2u∂φ2

=∂2u∂r2 +

2r∂u∂r

+1

r2 sin2 θ

∂2u∂φ2 +

1r2

∂2u∂θ2 +

cot θr2

∂u∂θ

(86)

Another equivalent form is:

∆ spherical u =1r∂2

∂r2

(ru

)+

1r2 sin θ

∂

∂θ

(sin θ

∂u∂θ

)+

1r2 sin2 θ

∂2u∂φ2 (87)

Note that some authors interchange θ and φ so that needs to be kept in mind (see the figure at the start ofthis section). For instance, if this occurs the form will be as follows (based on (87) ):

∆ spherical u =1r∂2

∂r2

(ru

)+

1r2 sin φ

∂

∂φ

(sin φ

∂u∂φ

)+

1r2 sin2 φ

∂2u∂θ2 (88)

21

5 The Laplacian in cylindrical coordinates – brute force

We start with the basic coordinate relationships and list the relevant derivatives:

x =r cos φy =r sin φz =z

(89)

r =

√x2 + y2

φ = arctan( y

x

)z =z

(90)

As before we work out the relevant building block derivatives:

∂r∂x

=2x

2√

x2 + y2

=x√

x2 + y2

=xr

(91)

22

∂r∂y

=y√

x2 + y2

=yr

(92)

∂φ

∂x=

1(1 + ( y

x )2) × −y

x2

= −y

x2 + y2

= −yr2

(93)

∂φ

∂y=

1(1 + ( y

x )2) × 1

x

=x

x2 + y2

=xr2

(94)

Hence:

∂u∂x

=∂u∂r

∂r∂x

+∂u∂φ

∂φ

∂x+∂u∂z

∂z∂x︸︷︷︸=0

=∂u∂r

∂r∂x

+∂u∂φ

∂φ

∂x

=xr∂u∂r−

yr2

∂u∂φ

(95)

We now work out ∂2u∂x2 :

∂2u∂x2 =

∂

∂x

(∂u∂x

)=∂

∂x

( xr∂u∂r−

yr2

∂u∂φ

)=

xr∂

∂x

(∂u∂r

)︸ ︷︷ ︸

A

+∂u∂r

∂

∂x

( xr

)︸ ︷︷ ︸

B

−yr2

∂

∂x

(∂u∂φ

)︸ ︷︷ ︸

C

−∂u∂φ

∂

∂x

( yr2

)︸ ︷︷ ︸

D

(96)

23

A =xr∂

∂x

(∂u∂r

)=

xr

[ ∂∂r

(∂u∂r

) ∂r∂x

+∂

∂φ

(∂u∂r

) ∂φ∂x

]=

xr

[ xr∂2u∂r2 −

yr2

∂2u∂φ ∂r

]=

x2

r2

∂2u∂r2 −

xyr3

∂2u∂φ ∂r

(97)

B =∂u∂r

∂

∂x

( xr

)=∂u∂r

∂

∂x

( x√x2 + y2

)=∂u∂r

( y2

(x2 + y2)32

)=

y2

r3

∂u∂r

(98)

C =−yr2

∂

∂x

(∂u∂φ

)=−yr2

[ ∂∂r

(∂u∂φ

) ∂r∂x

+∂

∂φ

(∂u∂φ

) ∂φ∂x

]=−yr2

[ xr∂2u∂r ∂φ

−yr2

∂2u∂φ2

]=−xyr3

∂2u∂r ∂φ

+y2

r4

∂2u∂φ2

(99)

D = −∂u∂φ

∂

∂x

( yr2

)= −

∂u∂φ

[ ∂∂r

( yr2

) ∂r∂x

+∂

∂φ

( yr2

)︸ ︷︷ ︸

=0

∂φ

∂x

]

= −∂u∂φ

[ xr−2yr3

]=

2xyr4

∂u∂φ

(100)

Therefore:

24

∂2u∂x2 =A + B + C + D

=x2

r2

∂2u∂r2 −

xyr3

∂2u∂φ ∂r

+y2

r3

∂u∂r−

xyr3

∂2u∂r∂φ

+y2

r4

∂2u∂φ2

+2xyr4

∂u∂φ

=x2

r2

∂2u∂r2 −

2xyr3

∂2u∂φ ∂r

+y2

r3

∂u∂r

+y2

r4

∂2u∂φ2 +

2xyr4

∂u∂φ

(101)

We now repeat the process for ∂2u∂y2 :

∂u∂y

=∂u∂r

∂r∂y

+∂u∂φ

∂φ

∂y+∂u∂z

∂z∂y︸︷︷︸=0

=∂u∂r

∂r∂y

+∂u∂φ

∂φ

∂y

=y√

x2 + y2

∂u∂r

+x

x2 + y2

∂u∂φ

(102)

∂2u∂y2 =

∂

∂y

(∂u∂y

)=∂

∂y

( y√x2 + y2

∂u∂r

+x

x2 + y2

∂u∂φ

)=

y√x2 + y2

∂

∂y

(∂u∂r

)︸ ︷︷ ︸

E

+∂u∂r

∂

∂y

( y√x2 + y2

)︸ ︷︷ ︸

F

+x

x2 + y2

∂

∂y

(∂u∂φ

)︸ ︷︷ ︸

G

+∂u∂φ

∂

∂y

( xx2 + y2

)︸ ︷︷ ︸

H

(103)

E =y√

x2 + y2

∂

∂y

(∂u∂r

)=

y√x2 + y2

[ ∂∂r

(∂u∂r

)∂r∂y

+∂

∂φ

(∂u∂r

)∂φ∂y

+∂

∂z

(∂u∂r

) ∂z∂y︸︷︷︸=0

]

=y√

x2 + y2

( y√x2 + y2

∂2u∂r2 +

xx2 + y2

∂2u∂φ ∂r

)=

y2

r2

∂2u∂r2 +

xyr3

∂2u∂φ ∂r

(104)

25

F =∂u∂r

∂

∂y

( y√x2 + y2

)=∂u∂r

( x2 + y2 − y2

r3

)=

x2

r3

∂u∂r

(105)

G =x

x2 + y2

∂

∂y

(∂u∂φ

)=

xx2 + y2

[ ∂∂r

(∂u∂φ

) ∂r∂y

+∂

∂φ

(∂u∂φ

) ∂φ∂y

+∂

∂z

(∂u∂φ

) ∂z∂y︸︷︷︸=0

]

=x

x2 + y2

[ y√x2 + y2

∂2u∂r ∂φ

+x

x2 + y2

∂2u∂φ2

]=

xyr3

∂2u∂r ∂φ

+x2

r4

∂2u∂φ2

(106)

H =∂u∂φ

∂

∂y

( xx2 + y2

)=∂u∂φ

−2xy(x2 + y2)2

=−2xy

r4

∂u∂φ

(107)

Therefore:

∂2u∂y2 =E + F + G + H

=y2

r2

∂2u∂r2 +

xyr3

∂2u∂φ ∂r

+x2

r3

∂u∂r

+xyr3

∂2u∂r ∂φ

+x2

r4

∂2u∂φ2

−2xyr4

∂u∂φ

=y2

r2

∂2u∂r2 +

2xyr3

∂2u∂φ ∂r

+x2

r3

∂u∂r

+x2

r4

∂2u∂φ2 −

2xyr4

∂u∂φ

(108)

Finally, we have:

26

∆ cylindrical =∂2u∂x2 +

∂2u∂y2 +

∂2u∂z2

=( x2 + y2

r2

) ∂2u∂r2 +

(2xyr3 −

2xyr3

) ∂2u∂r ∂φ

+(y2

r4 +x2

r4

) ∂2u∂φ2

+(y2

r3 +x2

r3

) ∂u∂r

+(2xy

r4 −2xyr4

) ∂u∂φ

=∂2u∂r2 +

1r2

∂2u∂φ2 +

1r∂u∂r

+∂2u∂z2

(109)

5.1 Rotational invariance of the Laplacian in two dimensions

Rotational invariance is a fundamental property of the Laplacian and in 2 dimensions one can demonstrateit as follows.

The transformation equations are as follows:

x′

=x cos θ + y sin θ

y′

= − x sin θ + y cos θ(110)

In matrix form this is:

(x′

y′

)=

(cos θ sin θ− sin θ cos θ

) (xy

)(111)

Having got this far there are no horrors of principle in what follows. What we want to show is this:

∂2u∂x2 +

∂2u∂y2 =

∂2u∂x′2

+∂2u∂y′2

(112)

As before we start off with ∂u∂x and ∂u

∂y in terms of x′

and y′

:

27

∂u∂x

=∂u∂x′

∂x′

∂x+∂u∂y′

∂y′

∂x

= cos θ∂u∂x′− sin θ

∂u∂y′

=Ux

(113)

∂u∂y

=∂u∂x′

∂x′

∂y+∂u∂y′

∂y′

∂y

= sin θ∂u∂x′

+ cos θ∂u∂y′

=Uy

(114)

Hence we have:

∂2u∂x2 =

∂Ux

∂x

=∂Ux

∂x′∂x

′

∂x+∂Ux

∂y′∂y′

∂x

= cos θ∂Ux

∂x′− sin θ

∂Ux

∂y′

= cos θ∂

∂x′(

cos θ∂u∂x′− sin θ

∂u∂y′

)− sin θ

∂

∂y′(

cos θ∂u∂x′− sin θ

∂u∂y′

)= cos2 θ

∂2u∂x′2

+ sin2 θ∂2u∂y′2− sin θ cos θ

∂2u∂x′ ∂y′

− sin θ cos θ∂2u

∂y′ ∂x′

(115)

∂2u∂y2 =

∂Uy

∂y

=∂Uy

∂x′∂x

′

∂y+∂Uy

∂y′∂y′

∂y

= sin θ∂Uy

∂x′+ cos θ

∂Uy

∂y′

= sin θ∂

∂x′(

sin θ∂u∂x′

+ cos θ∂u∂y′

)+ cos θ

∂

∂y′(

sin θ∂u∂x′

+ cos θ∂u∂y′

)= sin2 θ

∂2u∂x′2

+ cos2 θ∂2u∂y′2

+ sin θ cos θ∂2u

∂x′ ∂y′+ sin θ cos θ

∂2u∂y′ ∂x′

(116)

Thus:∂2u∂x2 +

∂2u∂y2 = (sin2 θ + cos2 θ)

( ∂2u∂x′2

+∂2u∂y′2

)=∂2u∂x′2

+∂2u∂y′2

(117)

28

What we have just shown is that the Laplacian is invariant under a 2-dimensional orthogonal transforma-

tion x′ = P(θ) x where P(θ) =

(cos θ sin θ− sin θ cos θ

)Note that if we had combined a translation with the rotation so that the transformation equations wereof the following form, where a, b are constants, the invariance still holds because the derivatives of theconstants are zero:

x′

=x cos θ + y sin θ + a

y′

= − x sin θ + y cos θ + b(118)

Now to show that the Laplacian is rotationally invariant under a 3-dimensional orthogonal rotation onecan somewhat laboriously replicate the 2-dimensional logic. Thus one could start with y = P x where Pis a 3 × 3 orthogonal matrix where P = [pi j]. Then:

yi =

n∑j=1

pi j x j (119)

∂yi

∂x j= pi j (120)

Then:

∂ u(Px)∂x1

=

3∑k=1

∂ u(y)∂yk

∂yk

∂y1

=

3∑k=1

pk 1∂ u(y)∂yk

(121)

After some labour we find that:

∂2 u(Px)∂x1

2 =p211∂2u(y)∂y1

2 + p11 p21∂2u(y)∂y1 ∂y2

+ p11 p13∂2u(y)∂y1 ∂y3

+ p21 p11∂2u(y)∂y2 ∂y1

+ p221∂2u(y)∂y2

2 + p21 p31∂2u(y)∂y2 ∂y3

+ p31 p11∂2u(y)∂y3 ∂y1

+ p31 p21∂2u(y)∂y2 ∂y3

+ p231∂2u(y)∂y3

2

(122)

If we keep going we ultimately get this:

(p211 + p2

12 + p213)︸ ︷︷ ︸

=1

∂2u(y)∂y1

2 + (p221 + p2

22 + p223)︸ ︷︷ ︸

=1

∂2u(y)∂y2

2 + (p231 + p2

32 + p233)︸ ︷︷ ︸

=1

∂2u(y)∂y3

2 (123)

29

The orthogonality of P ensures that the coefficients in (123) are 1. We can however get this result morecompactly as follows by using matrix notation throughout. First of all, because P = [pi j] is an orthogonalmatrix we have:

PPT = I (124)

The (i, j) element of PPT is∑3

k=1 pik p jk and because I is the identity matrix we must have:

3∑k=1

pik p jk = δi j =

1 if i = j0 if i , j (125)

Using matrix notation we can show the rotational invariance of the Laplacian in a few steps as follows.As before P is an n × n orthogonal matrix such that PPT = I. For x ∈ R3 we have:

v(x) = u(Px) (126)

What we want to show is that ∆u = 0 implies that ∆v = 0. The (i, j) element of P is pi j. The partialderivative operator we write as Di = ∂

∂xior Di = ∂

∂yias the context requires. Note that if v(x) = u(Px) then

the jth element of Px is y j =∑3

k=1 p jk xk. So ∂y j

∂xi= p ji

Thus we have:

Di v(x) =

3∑k=1

Dku(Px)pki (127)

Therefore:

Di j v(x) =

3∑l=1

3∑k=1

Dklu(Px)pki pl j (128)

Hence:

∆v(x) =

3∑i=1

Dii

=

3∑i=1

3∑l=1

3∑k=1

Dklu(Px)pki pli

=

3∑l=1

3∑k=1

Dklu(Px)( 3∑

i=1

pki pli

)=

3∑l=1

3∑k=1

Dklu(Px) δkl

=∆u(Px)=0

(129)

30

Note that the above line of argument generalises to d dimensions. One of the reasons that the Laplacian isso ubiquitous in physics is that it can be used to express physical laws which do not depend on a special orpreferred position. The Laplace operator also has spherical symmetry so that solutions which only dependon radial displacement are invariant under rotations about a chosen point ξ. Let u = ψ(r) where:

r = |x − ξ| =

√√√ 3∑i=1

(xi − ξi)2 (130)

We want to work out ∆u for this type of solution.

∂u∂xi

=dψdr

∂r∂xi

=ψ′

(r)xi − ξi

r

(131)

∂2u∂x2

i

=∂

∂xi

(ψ′

(r)xi − ξi

r

)=

xi − ξi

r∂

∂r

(ψ′

(r)xi − ξi

r

)+ψ′

(r)r

=(xi − ξi)2

r

[ rψ′′

(r) − ψ′

(r)r2

]+ψ′

(r)r

(132)

Hence:

∆u =

3∑i=1

[(xi − ξi)2

r

( rψ′′

(r) − ψ′

(r)r2

)+ψ′

(r)r

]=

r2

r

( rψ′′

(r) − ψ′

(r)r2

)+ 3

ψ′

(r)r

=ψ′′

(r) + 2ψ′

(r)r

=0

(133)

In relation to (132) note that if w = g(r, xi) then:

∂w∂xi

=∂g∂r

∂r∂xi

+∂g∂xi

∂xi

∂xi=∂g∂r

∂r∂xi

+∂g∂xi

(134)

For n dimensions the formula follows immediately from (133) to be as follows:

∆u = ψ′′

(r) +n − 1

rψ′

(r) = 0 (135)

31

To solve ψ′′

(r) + n−1r ψ

′

= 0 we seek a solution of the form ψ′

(r) = Crk for C a non-zero constant.Then:

ψ′′

(r) =Ckrk−1

∴ Ckrk−1 +n − 1

rCrk = 0

(136)

which implies that k = 1 − n.

Integrating ψ′

(r) = Cr1−n we get ψ(r) = C2−n r2−n if n > 2 or ψ(r) = C ln r. For instance when n = 3 we get

ψ(r) = Cr .

For more detail see [5], Chapter 4.

32

6 Derivation of the Laplacian in cylindrical coordinates using gradi-ent and divergence techniques

The high level recipe for this line of approach comprises three steps (see [8]) :

1. Obtain an expression for the divergence of F ie:

div F =1r∂

∂r(rFr) +

1r∂Fθ

∂θ+∂Fz

∂z(137)

where Fr, Fθ, Fz are the components of F in the drections of unit vectors er, eθ, ez respectively.

To do this we start with this definition of the divergence:

33

div F = lim∆V→0

1∆V

"F n dS (138)

where the volume limit is taken about a point (x, y, z) and n is a unit normal to the surface. Thus we needto look at an elementary volume in cylindrical coordinates and then calculate the relevant surface integralfor each face (see the diagram above which has the unit normals indicated).

2. Obtain an expression for the gradient in cylindrical coordinates. To do this we start with a Taylorexpansion of a scalar function f (r, θ, z). Thus the change in f as we move from (r, θ, z) to (r + ∆r, θ +

∆θ, z + ∆z) is:

∆ f =∂ f∂r

∆r +∂θ

∂θ∆θ +

∂ f∂z

∆z + higher order terms (139)

Then we have to get an expression for ∆ f in terms of the vectorial displacement ∆s. Recall that inCartesian coordinates we will have:

∆s = ∆x er + ∆y eθ + ∆z ez (140)

where er, eθ, ez are the relevant unit basis vectors.

so that:

∆ f =∂ f∂r

∆r +∂θ

∂θ∆θ +

∂ f∂z

∆z + higher order terms

=(∂ f∂x

er +∂ f∂y

eθ +∂ f∂z

ez) ∆s

(141)

But ∆s = u ∆s where u is a unit vector in the direction of ∆s and ∆s = |∆s|.

Therefore:

lim∆s→0

∆ f∆s

=d fds

=(∂ f∂x

er +∂ f∂y

eθ +∂ f∂z

ez) u

=(∇ f ) u(142)

3. Finally, the Laplacian is:

∆ f = ∇2 f = ∇ (∇ f ) (143)

We start with an elementary volume ∆V in cylindrical coordinates. This is a curved wedge of height ∆z,radial length ∆r and angular arc length (r + ∆r)∆θ = r∆θ + ∆r∆θ = r∆θ to first order (see the abovediagram). Hence:

∆V = r∆r ∆θ∆z (144)

34

If S i represents a surface (see above diagram ) with outward normal n then we have the following,assuming a wedge centred at (r, θ, z).

"S 1

F n dS =

"S 1

Fr dS

≈Fr

(r +

∆r2, θ, z

)(r +

∆r2

)∆θ∆z

(145)

"S 2

F n dS = −

"S 2

Fr dS

≈ − Fr

(r −

∆r2, θ, z

)(r −

∆r2

)∆θ∆z

(146)

Therefore:

1∆V

"S 1+S 2

F n dS =1

r∆r∆θ∆z

[Fr

(r +

∆r2, θ, z

)(r +

∆r2

)− Fr

(r −

∆r2, θ, z

)(r −

∆r2

)]∆θ∆z

=1

r∆r

[Fr

(r +

∆r2, θ, z

)(r +

∆r2

)− Fr

(r −

∆r2, θ, z

)(r −

∆r2

)] (147)

Therefore as ∆V → 0, ∆r → 0:

lim∆V→0

1∆V

"S 1+S 2

F n dS =1r∂

∂r

(rFr

)(148)

We now replicate the same process for surfaces S 3 and S 4.

"S 3

F n dS =

"S 3

Fz dS

≈Fz

(r, θ, z +

∆z2

)r∆r∆θ

(149)

"S 4

F n dS = −

"S 4

Fz dS

≈ − Fz

(r, θ, z −

∆z2

)r∆r∆θ

(150)

Therefore:

1∆V

"S 3+S 4

F n dS =1

r∆r∆θ∆z

[Fz

(r, θ, z +

∆z2

)− Fz

(r, θ, z −

∆z2

)]r∆r∆θ

=1∆z

[Fz

(r, θ, z +

∆z2

)− Fz

(r, θ, z −

∆z2

)] (151)

35

Hence:

lim∆V→0

1∆V

"S 3+S 4

F n dS =∂Fz

∂z(152)

Finally we do the calculation for S 5 and S 6.

"S 6

F n dS =

"S 6

Fθ dS

≈Fθ

(r, θ +

∆θ

2, z

)∆r∆z

(153)

"S 5

F n dS = −

"S 5

Fθ dS

≈ − Fθ

(r, θ −

∆θ

2, z

)∆r∆z

(154)

Therefore:

lim∆V→0

1∆V

"S 5+S 6

F n dS =≈1

r∆r∆θ∆z

[Fθ

(r, θ +

∆θ

2, z

)− Fθ

(r, θ −

∆θ

2, z

)]∆r∆z

=1

r∆θ

[Fθ

(r, θ +

∆θ

2, z

)− Fθ

(r, θ −

∆θ

2, z

)] (155)

Therefore:

lim∆V→0

1∆V

"S 5+S 6

F n dS =1r∂Fθ

∂θ(156)

So putting it all together we have:

div F =1r∂

∂r

(rFr

)+

1r∂Fθ

∂θ+∂Fz

∂z(157)

In accordance with the recipe we now need to find the gradient in cylindrical coordinates. If f (r, θ, z) weapply Taylor’s theorem to obtain the change in f due to a displacement from (r, θ, z) to (r + ∆r, θ+ ∆θ, z +

∆z):

∆ f =∂ f∂r

∆r +∂ f∂θ

∆θ +∂ f∂z

∆z + higher order terms (158)

To write ∆ f in terms of the vectorial displacement ∆s we need to express ∆s in terms of its natural basisas follows:

36

∆s = er∆r + eθr∆θ + ez∆z (159)

In relatiion to (159), recall that the displacement in the direction of increasing θ is (r+∆r)∆θ = r∆θ+∆r∆θand we ignore the second order term ∆r∆θ giving r∆θ.

So:

∆ f =(er∂ f∂r

+ eθ1r∂ f∂θ

+ ez∂ f∂z

) ∆~s + higher order terms (160)

Note the factor of 1r which ensures on doing the dot product we get the right form of ∆ f .

But ∆s = u ∆s where u is a unit vector in the direction of s.

Therefore:

lim∆s→0

∆ f∆s

=d fds

=(er∂ f∂r

+ eθ1r∂ f∂θ

+ ez∂ f∂z

)︸ ︷︷ ︸

∇ f

u (161)

Thus:

∇cyl =

∂∂r

1r∂∂θ

∂∂z

(162)

We now have all the ingredients to finish the recipe by calculating the Laplacian as follows (see (143)):

∆Φ = ∇2Φ = ∇ (∇Φ) = ∇ F = div F (163)

where ∇cylΦ = F

i.e: ∂Φ∂r

1r∂Φ∂θ

∂Φ∂z

=

Fr

Fθ

Fz

(164)

Therefore using (124) and (164):

∆Φ =1r∂

∂r

(r∂Φ

∂r

)+

1r2

∂

∂θ

(∂Φ

∂θ

)+∂

∂z

(∂Φ

∂z

)=

1r∂

∂r

(r∂Φ

∂r

)+

1r2

∂2Φ

∂θ2 +∂2Φ

∂z2

(165)

37

7 Derivation of the Laplacian in spherical coordinates using gradientand divergence techniques

Using the same recipe we now calculate the Laplacian in spherical coordinates. This is more intricatebecause one has to be careful with the form of each elementary surface. In this representation (r, θ, φ)note that some authors interchange the roles of θ and φ so that needs to be kept in mind when comparingthe final form of the Laplacian.

The elementary volume is made up of radial and two angular elementary arcs as follows. To obtain thecircular arc which is parallel to the (x, y) plane (ie in the φ direction) the radius is (r + ∆r) sin θ so that thearc length is (r + ∆r) sin θ∆φ = r sin θ∆φ + ∆r∆φ sin θ︸ ︷︷ ︸

ignore as second order

≈ r sin θ∆φ. The radial dimension of the

elementary volume has length ∆r and the arc length in the θ direction is:

38

(r + ∆r)∆θ = r∆θ + ∆r∆θ︸︷︷︸ignore as second order

≈ r∆θ. Thus:

∆V = r sin θ∆φ × ∆r × r∆θ = r2 sin θ∆r ∆φ∆θ (166)

For convenience the elementary surface areas are set out below. Each elementary surface is centred on(r, θ, φ) and is used in working out the approximations to the relevant surface integrals:

∆S 1 =(r +∆r2

) sin θ∆φ × (r +∆r2

)∆θ = (r +∆r2

)2 sin θ∆φ∆θ

∆S 2 =(r −∆r2

) ∆θ × (r −∆r2

) sin θ∆φ = (r −∆r2

)2 sin θ∆φ∆θ

∆S 3 =(r +∆r2

) sin(θ +∆θ

2) ∆r ∆φ

∆S 4 =(r +∆r2

) sin θ∆r ∆φ

∆S 5 =(r +∆r2

) ∆r ∆θ

∆S 6 =(r +∆r2

) ∆r ∆θ

(167)

"S 1

F n dS =

"S 1

Fr dS

≈Fr

(r +

∆r2, θ, φ

)∆S 1

=Fr

(r +

∆r2, θ, φ

)(r +

∆r2

)2 sin θ∆φ∆θ

(168)

"S 2

F n dS = −

"S 2

Fr dS

≈ − Fr

(r −

∆r2, θ, φ

)∆S 2

= − Fr

(r −

∆r2, θ, φ

)(r −

∆r2

)2 sin θ∆φ∆θ

(169)

Therefore:

lim∆V→0

1∆V

"S 1+S 2

F n dS ≈1

r2 sin θ∆r∆φ∆θ

[Fr

(r +

∆r2, θ, φ

)(r +

∆r2

)2 sin θ∆φ∆θ−

Fr

(r −

∆r2, θ, φ

)(r −

∆r2

)2 sin θ∆φ∆θ

]

= lim∆r→0

1r2

[Fr

(r + ∆r

2 , θ, φ)

(r + ∆r2 )2 − Fr

(r − ∆r

2 , θ, φ)

(r − ∆r2 )2

∆r

]→

1r2

∂

∂r

(r2Fr

)(170)

39

"S 3

F n dS =

"S 3

Fθ dS

≈Fθ

(r, θ +

∆θ

2, φ

)∆S 3

=Fθ

(r, θ +

∆θ

2, φ

)(r +

∆r2

) sin(θ +∆θ

2) ∆r ∆φ

(171)

"S 4

F n dS = −

"S 4

Fθ dS

≈ − Fθ

(r, θ −

∆θ

2, φ

)∆S 4

= − Fθ

(r, θ −

∆θ

2, φ

)(r +

∆r2

) sin θ∆r ∆φ

(172)

lim∆V→0

1∆V

"S 3+S 4

F n dS

≈1

r2 sin θ∆r∆φ∆θ

[Fθ

(r, θ +

∆θ

2, φ

)(r +

∆r2

) sin(θ +∆θ

2) ∆r ∆φ

−Fθ

(r, θ −

∆θ

2, φ

)(r +

∆r2

) sin θ∆r ∆φ

]

= lim∆r→0

1r2 sin θ

[Fθ

(r, θ + ∆θ

2 , φ)

(r + ∆r2 ) sin(θ + ∆θ

2 ) − Fθ

(r, θ − ∆θ

2 , φ)

(r + ∆r2 ) sin θ

∆θ

]→

1r sin θ

∂

∂θ

(sin θ Fθ

)(173)

"S 5

F n dS =

"S 5

Fφ dS

≈Fφ

(r, θ, φ +

∆φ

2

)∆S 5

=Fφ

(r, θ, φ +

∆φ

2

)(r +

∆r2

) ∆r ∆θ

(174)

"S 6

F n dS = −

"S 6

Fφ dS

≈ − Fφ

(r, θ, φ −

∆Φ

2

)∆S 6

= − Fφ

(r, θ, φ −

∆φ

2

)(r +

∆r2

) ∆r ∆θ

(175)

Therefore:

40

lim∆V→0

1∆V

"S 5+S 6

F n dS =1

r2 sin θ∆r∆φ∆θ

"S 5+S 6

Fφ dS

≈1

r2 sin θ∆r∆φ∆θ

"S 5+S 6

Fφ ∆S 6

=1

r2 sin θ∆r∆φ∆θ

[Fφ

(r, θ, φ +

∆φ

2

)(r +

∆r2

) ∆r ∆θ

−Fφ

(r, θ, φ −

∆φ

2

)(r +

∆r2

) ∆r ∆θ

]

=1

r2 sin θ

[Fφ

(r, θ, φ +

∆φ2

)(r + ∆r

2 ) − Fφ

(r, θ, φ − ∆φ

2

)(r + ∆r

2 )

∆φ

]→

1r sin θ

∂Fφ

∂φ

(176)

Putting it all together we have:

div F =1r2

∂

∂r

(r2 Fr

)+

1r sin θ

∂

∂θ

(sin θ Fθ

)+

1r sin θ

∂Fφ

∂φ(177)

As before we need to work ou the gradient and as a first step we need to determine the elementary vectorialdisplacement ∆s. The displacements in the 3 directions are (see the diagram above):

radial direction: ∆rθ direction: (r + ∆r)∆θ = r∆θ + ∆r∆θ︸︷︷︸

second order

≈ r∆θ

φ direction: (r + ∆r) sin θ∆φ = r sin θ∆φ + sin θ∆r∆φ︸ ︷︷ ︸second order

≈ r sin θ∆φ

Thus in terms of its natural basis the vectorial displacement is:

∆s = er ∆r + eθ r∆θ + eφ r sin θ∆φ (178)

We also have:

∆ f =∂ f∂r

∆r +∂θ

∂θ∆θ +

∂ f∂φ

∆φ + higher order terms (179)

Thus we can write:

∆ f =(er∂ f∂r

+ eθ1r∂ f∂θ

+ eφ1

r sin θ∂ f∂φ

) ∆s + higher order terms (180)

But ∆s = u ∆s where u is a unit vector in the direction of s.

Therefore:

41

lim∆s→0

∆ f∆s

=d fds

=(er∂ f∂r

+ eθ1r∂ f∂θ

+ eφ1

r sin θ∂ f∂φ

)︸ ︷︷ ︸

∇ f

u (181)

Thus:

∇sph =

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

(182)

As before we now have all the ingredients to finish the recipe by calculating the Laplacian as follows (Iuse Φ as the function symbol while φ is the angular variable ):

∆Φ = ∇2Φ = ∇ (∇Φ) = ∇ F = div F (183)

where ∇sphΦ = F

i.e:

∂Φ∂r

1r∂Φ∂θ

1r sin θ

∂Φ∂φ

=

Fr

Fθ

Fφ

(184)

Therefore using (177):

∆Φ =1r2

∂

∂r

(r2 ∂Φ

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂Φ

∂θ

)+

1r2 sin2 θ

∂2Φ

∂φ2 (185)

Note that if θ and φ are interchanged (as some authors do ) the form of the Laplacian will be:

∆Φ =1r2

∂

∂r

(r2 ∂Φ

∂r

)+

1r2 sin φ

∂

∂φ

(sin φ

∂Φ

∂φ

)+

1r2 sin2 φ

∂2Φ

∂θ2 (186)

The Laplacian (185) can be written in the alternative form:

∆Φ =1r∂2

∂r2

(rΦ

)+

1r2 sin θ

∂

∂θ

(sin θ

∂Φ

∂θ

)+

1r2 sin2 θ

∂2Φ

∂φ2 (187)

That the two forms are equivalent is based on this equality:

1r2

∂

∂r

(r2 ∂Φ

∂r

)=

1r∂2

∂r2

(rΦ

)(188)

42

We can get the gradient in spherical coordinates quite efficiently using the following approach. In the(r, θ, φ) coordinates a general point x has the following representation. See the diagram at the beginningof this section:

x =

r sin θ cos φr sin θ sin φ

r cos θ

(189)

Then:

dx =

sin θ cos φsin θ sin φ

cos θ

dr + r

cos θ cos φcos θ sin φ− sin θ

dθ + r

− sin θ sin φsin θ cos φ

0

dφ (190)

We can write (190) as follows in terms of unit vectors er, eθ, eφ:

dx = er dr + r eθ dθ + r sin θ eφ dφ (191)

where:

er =

sin θ cos φsin θ sin φ

cos θ

(192)

eθ =

cos θ cos φcos θ sin φ− sin θ

(193)

eφ =

− sin φcos φ

0

(194)

It is easily verified that er, eθ, eφ are unit vectors eg |er| =

√sin2 θ cos2 φ + sin2 θ sin2 φ + cos2 θ =√

sin2 θ + cos2 θ = 1.

The gradient of Φ is:

∇Φ =∂Φ

∂rdr +

∂Φ

∂θdθ +

∂Φ

∂φdφ (195)

But:

dΦ = ∇Φ dx (196)

So using (191) we have:

43

∂Φ

∂rdr +

∂Φ

∂θdθ +

∂Φ

∂φdφ =∇Φ

(er dr + r eθ dθ + r sin θ eφ dφ

)=∇Φ er dr + ∇Φ eθ r dθ + ∇Φ eφ r sin θ dφ

(197)

Equating coefficients we have:

∂Φ

∂r=∇Φ er

1r∂Φ

∂θ=∇Φ eθ

1r sin θ

∂Φ

∂φ=∇Φ eφ

(198)

The LHS of (198) comprise the projections of the vector ∇Φ onto the respective unit vectors. There-fore:

∇Φ =(∇Φ er) er + (∇Φ eθ) eθ + (∇Φ eφ) eφ

=∂Φ

∂rer +

1r∂Φ

∂θeθ +

1r sin θ

∂Φ

∂φeφ

(199)

Thus, as before, the gradient operator has the following form in spherical coordinates:

∇sph =

∂∂r

1r∂∂θ

1r sin θ

∂∂φ

(200)

One can apply the same logic to the gradient in cylindrical coordinates.

8 A more general approach to the Laplacian

The final approach I will look at revolves around a general expression for the Laplacian in orthogonalcurvilinear coordinates. This is the most powerful method because all the “usual suspects” (Cartesian,polar, cylindrical, spherical etc) are all contained within the general formula. Even more generally thisprocedure can comprehend non-orthogonal systems, although I will not deal with that possibility. Thedownside is that there is some differential geometry and if you serious, tensor theory.

So let’s begin. We assume we have an orthogonal curvilinear coordinate system. If you have surfacesu1 = c1, u2 = c2 and u3 = c3 where the ci are constants, each pair of the surfaces will intersect in acurvilinear coordinate curve and if they intersect at right angles we will have an orthogonal system.

We now have to relate the position vector of a point in Cartesian coordinates to these new coordinates.The relationship is something of this form:

44

x = x(u1, u2, u3)y = y(u1, u2, u3)z = z(u1, u2, u3)

(201)

For instance in cylindrical coordinates (r, φ, z) we have:

x = r cos φy = r sin φ

z = z(202)

The position vector of a point P is :

r = xi + yj + zk (203)

i, j,k are the standard Cartesian unit vectors.

The crux of this approach is to obtain a representation of position vectors in terms of tangential (con-travariant) unit base vectors and normal (covariant) unit base vectors. Thus the relevant vector is repre-sented in either tangential or normal components. Once this is understood the mechanics are relativelystraightforward.

A tangent vector to u1 at P is simply:

∂r∂u1

(204)

and a unit vector in that direction is just:

e1 =

∂r∂u1∣∣∣ ∂r∂u1

∣∣∣ (205)

So we can write the tangent vector in terms of a scale factor h1 as follows:

∂r∂u1

= h1e1 (206)

where h1 =∣∣∣ ∂r∂u1

∣∣∣Replicating that logic we have for i = 1, 2, 3:

∂r∂ui

= hiei (207)

where hi =∣∣∣ ∂r∂ui

∣∣∣Now for the normal representation. We know from basic calculus that a vector which is normal to thesurface u1 = c1 at P is:

45

∇u1 (208)

Hence a unit vector in this direction is:

E1 =∇u1∣∣∣∇u1

∣∣∣ (209)

As before we have for i = 1, 2, 3:

Ei =∇ui∣∣∣∇ui

∣∣∣ (210)

The ∂r∂ui

and ∇ui are a reciprocal system of vectors in the sense that:

∂r∂ui· ∇u j =

1 if i = j

0 if i , j(211)

This is seen as follows: Since:

dr =∂r∂u1

du1 +∂r∂u2

du2 +∂r∂u3

du3 =

3∑i=1

∂r∂ui

dui (212)

when we take the dot product with ∇u1 (which is orthogonal to ∂r∂u1

) we get:

∇u1 · dr = du1 =

3∑i=1

(∇ui ·

∂r∂ui

)dui (213)

Hence:

∇u1 ·∂r∂u1

= 1

∇u1 ·∂r∂u2

= 0

∇u1 ·∂r∂u3

= 0

(214)

Similarly for ∇u2 and ∇u3.

To get a general expression for the gradient in orthogonal curvilinear coordinates we proceed as follows.We want ∇Φ:

∇Φ =

3∑i=1

fi ei (215)

46

dr =

3∑i=1

∂r∂ui

dui

=

3∑i=1

hi ei dui using (181)

(216)

But the directional derivative is:

dΦ =∇Φ dr

=

3∑i=1

hi fi dui

=

3∑i=1

∂Φ

∂uidui

(217)

Equating coefficients in the last two lines of (191) we have for i = 1, 2, 3:

fi =1hi

∂Φ

∂ui(218)

Hence:

∇Φ =

3∑i=1

ei

hi

∂Φ

∂ui(219)

We can also write (219) in operator form:

∇ =

3∑i=1

ei

hi

∂

∂ui(220)

Note that in the Cartesian coordinate system h1 = h2 = h3 = 1 and so we can identify e1 as i , e2 as jand e3 as k. This also means that the vectors ei for i = 1, 2, 3 have the same cyclic behaviour as i, j,kie

e1 × e2 =e3

e2 × e3 =e1

e3 × e1 =e2

(221)

In (219) let Φ = ui then:

47

∇ui =ei

hi=⇒ |∇ui| =

|ei|

hi=

1hi

(222)

Because we want an expression for ∇2Φ = ∇ ∇Φ we need a general expression for ∇ A = div A. Itturns out (as is shown below) that:

∇ (A1e1) = ∇ (A1h2h3 ∇u2 × ∇u3

)(223)

with similar expressions for other components.

To get the cross product in (223) we use (221-222):

∇u2 × ∇u3 =e2 × e3

h2 3=

e1

h2 h3(224)

Therefore:

e1 = h2h3 ∇u2 × ∇u3 (225)

Similiar expressions follow for e2 and e3 ie

h3h1 ∇u3 × ∇u1 =e2

h1h2 ∇u1 × ∇u2 =e3(226)

The basic result we need is this. For a vector B and differentiable scalar function Φ the followingholds:

∇ (ΦB

)=

(∇Φ

) B + Φ

(∇ B

)(227)

(227) is proved in the Appendix.

Now applying (227) to (223) with Φ = A1h2h3 and B = ∇u2 × ∇u3 we have:

∇ (A1e1) =∇ (A1h2h3 ∇u2 × ∇u3

)=∇

(A1h2h3

) ∇u2 × ∇u3 + A1h2h3 ∇

(∇u2 × ∇u3

)︸ ︷︷ ︸=0

=∇(A1h2h3

)

e2

h2×

e3

h3

=∇(A1h2h3

)

e1

h2h3

=[ e1

h1

∂

∂u1(A1h2h3) +

e2

h2

∂

∂u2(A1h2h3) +

e3

h3

∂

∂u3(A1h2h3)

]

e1

h2h3using (220)

=1

h1h2h3

∂

∂u1(A1h2h3)

(228)

48

Note that ∇ (∇u2 × ∇u3

)= 0 because of the following:

∇ (∇u2 × ∇u3

)=∇

e1

h2h3

=( 3∑

i=1

ei

hi

∂

∂ui

)

e1

h2h3

=0

(229)

We now repeat the process for the other factors:

∇ (A2e2) =∇ (A2h3h1 ∇u3 × ∇u1

)=∇

(A2h3h1

) ∇u3 × ∇u1 + A2h3h1 ∇

(∇u3 × ∇u1

)︸ ︷︷ ︸=0

=∇(A2h3h1

)

e3

h3×

e1

h1

=∇(A2h3h1

)

e2

h3h1

=[ e1

h1

∂

∂u1(A2h3h1) +

e2

h2

∂

∂u2(A2h3h1) +

e3

h3

∂

∂u3(A2h3h1)

]

e2

h3h1using (220)

=1

h1h2h3

∂

∂u2(A2h3h1)

(230)

∇ (A3e3) =∇ (A3h1h2 ∇u1 × ∇u2

)=∇

(A3h1h2

) ∇u1 × ∇u2 + A3h1h2 ∇

(∇u1 × ∇u2

)︸ ︷︷ ︸=0

=∇(A3h1h2

)

e1

h1×

e2

h2

=∇(A3h1h2

)

e3

h1h2

=[ e1

h1

∂

∂u1(A3h1h2) +

e2

h2

∂

∂u2(A3h1h2) +

e3

h3

∂

∂u3(A3h1h2)

]

e3

h1h2using (220)

=1

h1h2h3

∂

∂u3(A3h1h2)

(231)

Putting it all together we see that:

∇ A =∇ (A1e1 + A2e2 + A3e3)

∇ A =1

h1h2h3

[ ∂

∂u1(A1h2h3) +

∂

∂u2(A2h3h1) +

∂

∂u3(A3h1h2)

] (232)

We are now in a position to get the general form of the Laplacian in orthogonal curvilinear coordinates.If we let A = ∇Φ then Ai = 1

hi

∂Φ∂ui

and so by (232):

49

∆ =∇ ∇Φ

=∇2Φ

=1

h1h2h3

[ ∂

∂u1

(h2h3

h1

∂Φ

∂u1

)+

∂

∂u2

(h3h1

h2

∂Φ

∂u2

)+

∂

∂u3

(h1h2

h3

∂Φ

∂u3

)] (233)

How do you remember the scale factors h1, h2, h3 ? The easiest way is to remember what the elementaryvalue components are in each system. The volume element in orthogonal curvilinear coordinates u1, u2, u3is dV = h1h2h3 du1du2du3. For instance in Cartesian coordinates the elementary volume is dx × dy × dzso h1 = h2 = h3 = 1.

In cylindrical coordinates the elementary volume is ( see the diagram associated with (144) ):

dV = (dr) (rdθ) (dz) (234)

Note here that it is critical to properly associate the relevant factors with their basis vectors ie h1du1 =

1 × dr, h2du2 = r × dθ and h3du3 = 1 × dz. Hence in this case h1 = 1, h2 = r and h3 = 1.

Plugging these values into (233) we get:

∆Φ =1r

[ ∂∂r

(r∂Φ

∂r

)+∂

∂θ

(1r∂Φ

∂θ

)+∂

∂z

(r∂Φ

∂z

)]=

1r∂

∂r

(r∂Φ

∂r

)+

1r2

∂2Φ

∂θ2 +∂2Φ

∂z2

(235)

which is thankfully the same as (165).

For spherical coordinates the elementary volume is (see (166) and the diagram and discussion associatedwith it):

dV = (dr) (rdθ) (r sin θ dφ) (236)

Hence h1 = 1, h2 = r, and h3 = r sin θ.

Plugging these values into (233) we get:

∆Φ =1

r2 sin θ

[ ∂∂r

(r2 sin θ

∂Φ

∂r

)+∂

∂θ

(sin θ

∂Φ

∂θ

)+

∂

∂φ

( 1sin θ

∂Φ

∂φ

)]=

1r2

∂

∂r

(r2 ∂Φ

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂Φ

∂θ

)+

1r2 sin2 θ

∂2Φ

∂φ2

(237)

which is thankfully the same as (185).

The structure of (233) is actually easy to remember once you note how the scale factors figure in eachderivative and it is easy to work out the elementary volumes by drawing a diagram in order to get the scalefactors. Note the cyclic symmetry of the factors in each of the partial derivatives. That is for i = 1, therelevant fraction in the derivative has h1 on the bottom and h2 h3 on top and for i = 2 it cycles 2→ 3→ 1and for i = 3 it cycles 3→ 1→ 2.

50

Can we use (233) to work out the Laplacian in polar coordinates (r, θ)? Yes we can as long as we realisethat in polar coordinates the ”volume” element is a ”degenerate” one - it is an area which we will still calldV:

dV = dr × r dθ (238)

So h1 = 1 and h2 = r. Our expression for the Laplacian therefore has only two terms:

∆Φ =1

h1h2

[ ∂

∂u1

(h2

h1

∂Φ

∂u1

)+

∂

∂u2

(h1

h2

∂Φ

∂u2

)](239)

So plugging the factors in we get:

∆Φ =1r

[ ∂∂r

(r∂Φ

∂r

)+∂

∂θ

(1r∂Φ

∂θ

)]=

1r

[r∂2Φ

∂r2 +∂Φ

∂r+

1r∂2Φ

∂θ2

]=∂2Φ

∂r2 +1r∂Φ

∂r+

1r2

∂2Φ

∂θ2

(240)

which is the same as (45).

9 The final generalisation - the tensor form of the Laplacian

The expression in (233) can be generalised even further using the techniques of tensor calculus. I willonly state what the Laplacian looks like first (see [5], page 231 ):

∆ = ∇2Φ =1√

g∂

∂xk

(√g gkr ∂Φ

∂xr

)(241)

To prove (241) is a major undertaking which involves knowing how to perform covariant differentiationand several other things. This is already a long paper and I am not going to double its length by a fullblown account of tensor calculus). In summary one starts with an expression for the divergence:

div Ap =1√

g∂

∂xk

(√g Ak) (242)

The gradient of Φ is grad Φ = ∇Φ = ∂Φ∂xr which is defined as a covariant derivative of Φ often written as

Φ,r. The rank 1 contravariant tensor associated with Φ,r is Ak = gkr ∂Φ∂xr . Then:

∇2Φ = div (∇Φ) = div(gkr ∂Φ

∂xr

)=

1√

g∂

∂xk

(√g gkr ∂Φ

∂xr

)(243)

51

There are some things of a general nature to note about (241). First the determinant of the metric tensor,that is g, takes the place of h1h2h3 which makes sense because of the following. If the derivative of theposition vector is dr = ∂r

∂u1du1 + ∂r

∂u2du2 + ∂r

∂u3du3 then:

ds2 =

3∑p=1

3∑q=1

gpq dupduq (244)

or simply gpq dupduq using the Einstein summation convention. Note that by ds2 we mean (ds)2 ratherthan some contravariant index.

Thus:

gpq =∂r∂up∂r∂uq

(245)

In orthogonal coordinate systems under consideration in this paper gpq = 0 when p , q. Recall that inorthogonal curvilinear coordinates the elementary volume element is:

dV = |(h1du1e1) (h2du2e2 × (h3du3e3)| = h1h2h3 du1du2du3 (246)

where the ek =∂r∂uk

1∣∣∣ ∂r∂uk

∣∣∣ . In the tensor representation the scale factors h1, h2, h3 are replaced by√

g which in

turn reflects the scale factors for the volume.

The other point to note that in (233) we have factors such as h2h3h1

which are replaced by√

g gkr where gkr

is the conjugate metric tensor to gkr. The relationship between the metric tensor and its conjugate boilsdown to this:

g jk gpk = δpj (247)

where δpj is the Kronecker delta ie δp

j = 1 if p = j and 0 if p , j. In matrix terms the conjugate metrictensor involves calcuating cofactors and this is why you get terms such as h2h3

h1since you are multiplying

a determinant of the metric tensor (ie√

g ) by a cofactor type of term.

9.1 Applying the tensor formula to cylindrical coordinates

For cylindrical coordinates we have:

x1 =r cos θ

x2 =r sin θ

x3 =z

(248)

Hence:

52

dx1 = − r sin θ dθ + cos θ dr

dx2 =r cos θ dθ + sin θ dr

dx3 =dz

(249)

The squared arc length is:

ds2 = dr2 + r2dθ2 + dz2 (250)

You can now read off the metric tensor components (knowing that the off diagonal ones are 0):

g11 =1

g22 =r2

g33 =1g12 =g21 = 0g23 =g32 = 0g31 =g13 = 0

(251)

In matrix form the tensor looks like this:

(gi j) =

1 0 00 r2 00 0 1

(252)

The determinant of (252) is g = r2 hence√

g = r. Note that because of (247) we have:

g11 =1

g11

g22 =1

g22

g33 =1

g33

(253)

In (241) we need the conjugate metric tensor gkr and this has a simple matrix form where the diagonalterms are inverted:

(gi j) =

1 0 00 1

r2 00 0 1

(254)

We now plug the relevant bits into (241) as follows:

53

∇2Φ =1√

g∂

∂xk

(√g gkr ∂Φ

∂xr

)=

1r

[ ∂∂r

(r.1∂Φ

∂r

)+∂

∂θ

(r.

1r2

∂Φ

∂θ

)+∂

∂z

(r.1∂Φ

∂z

) ]=

1r∂

∂r

(r∂Φ

∂r

)+

1r2

∂2Φ

∂θ2 +∂2Φ

∂z2

(255)

which is the same as (165).

If we wanted to calculate the conjugate metric tensor using cofactors the approach is as follows (see (252)):

g11 =cofactor of g11

g=

1r2

∣∣∣∣∣∣r2 00 1

∣∣∣∣∣∣ = 1 (256)

g22 =cofactor of g22

g=

1r2

∣∣∣∣∣∣1 00 1

∣∣∣∣∣∣ =1r2 (257)

g33 =cofactor of g33

g=

1r2

∣∣∣∣∣∣1 00 r2

∣∣∣∣∣∣ = 1 (258)

9.2 Applying the tensor formula to spherical coordinates

For spherical coordinates we have:

x1 =r sin θ cos φ

x2 =r sin θ sin φ

x3 =r cos θ

(259)

Hence:

dx1 = sin θ cos φ dr + r cos θ cos φ dθ − r sin θ sin φ dφ

dx2 = sin θ sin φ dr + r cos θ sin φ dθ + r sin θ cos φ dφ

dx3 = cos θ dr − r sin θ dθ

(260)

The squared arc length is:

ds2 = dr2 + r2dθ2 + r2 sin2 θ dφ2 (261)

You can now read off the metric tensor components (knowing that the off diagonal ones are 0):

54

g11 =1

g22 =r2

g33 =r2 sin2 θ

g12 =g21 = 0g23 =g32 = 0g31 =g13 = 0

(262)

In matrix form the tensor looks like this:

(gi j) =

1 0 00 r2 00 0 r2 sin2 θ

(263)

The determinant of (263) is g = r4 sin2 θ hence√

g = r2 sin θ. Note that because of (247) we have:

(gi j) =

1 0 00 1

r2 00 0 1

r2 sin2 θ

(264)

We now plug the relevant bits into (241):

∇2Φ =1√

g∂

∂xk

(√g gkr ∂Φ

∂xr

)=

1r2 sin θ

[ ∂∂r

(r2 sin θ.1

∂Φ

∂r

)+∂

∂θ

(r2 sin θ.

1r2

∂Φ

∂θ

)+

∂

∂φ

(r2 sin θ.

1r2 sin2 θ

∂Φ

∂φ

) ]=

1r2

∂

∂r

(r2 ∂Φ

∂r

)+

1r2 sin θ

∂

∂θ

(sin θ

∂Φ

∂θ

)+

1r2 sin2 θ

∂2Φ

∂φ2

(265)

which is the same as (185).

If we wanted to calculate the conjugate metric tensor using cofactors the approach is as follows (see (263)):

g11 =cofactor of g11

g=

1r4 sin2 θ

∣∣∣∣∣∣r2 00 r2 sin2 θ

∣∣∣∣∣∣ = 1 (266)

g22 =cofactor of g22

g=

1r4 sin2 θ

∣∣∣∣∣∣1 00 r2 sin2 θ

∣∣∣∣∣∣ =1r2 (267)

g33 =cofactor of g33

g=

1r4 sin2 θ

∣∣∣∣∣∣1 00 r2

∣∣∣∣∣∣ =1

r2 sin2 θ(268)

55

9.3 Applying the tensor formula to polar coordinates

For polar coordinates we have:

x1 =r cos θ

x2 =r sin θ(269)

Hence:

dx1 = cos θ dr − r sin θ dθ

dx2 = sin θ dr + r cos θ dθ(270)

The squared arc length is:

ds2 = dr2 + r2dθ2 (271)

You can now read off the metric tensor components (knowing that the off diagonal ones are 0):

g11 =1

g22 =r2

g12 =g21 = 0

(272)

(gi j) =

(1 00 r2

)(273)

The determinant of (273) is g = r2 hence√

g = r. Note that because of (247) we have:

(gi j) =

(1 00 1

r2

)(274)

We now plug the relevant bits into (241):

∇2Φ =1√

g∂

∂xk

(√g gkr ∂Φ

∂xr

)=

1r

[ ∂∂r

(r.1∂Φ

∂r

)+∂

∂θ

(r.

1r2

∂Φ

∂θ

) ]=

1r∂

∂r

(r∂Φ

∂r

)+

1r2

∂2Φ

∂θ2

=∂2Φ

∂r2 +1r∂Φ

∂r+

1r2

∂2Φ

∂θ2

(275)

which is the same as (45).

56

10 Appendix

10.1 Change of variables

If we suppose a change of variables as follows:

ξ = Btx ie ξr =∑n

i=1 bir xi for r = 1, 2, . . . , n we then have the following:

uxi =∂u∂xi

=

n∑r=1

∂u∂ξr

∂ξr

∂xi

=

n∑r=1

bir∂u∂ξr

uxi x j =∂

∂x j

( ∂u∂xi

)=

n∑s=1

(uxi )ξs

∂ξs

∂x j

=

n∑s=1

b js(uxi )ξs

=

n∑s=1

b js∂

∂ξs

( n∑r=1

bir∂u∂ξr

)=

n∑r,s=1

birb js∂2u∂ξr∂ξs

(276)

Equation (6) was as follows and is transformed in terms of the ξi as follows:

n∑i, j=1

ai juxi x j +

n∑i, j=1

biuxi + cu = d (277)

So we have:

n∑r,s=1

( n∑i, j=1

birai jb js

)︸ ︷︷ ︸

crs

uξrξs +

n∑r=1

( n∑i=1

birbi

)uξr + cu = d (278)

10.2 Proof of equation (227)

We have to prove:

∇ (ΦB

)=

(∇Φ

) B + Φ

(∇ B

)(279)

57

We have from (194):

∇ =e1

h1

∂

∂u1+

e2

h2

∂

∂u2+

e3

h3

∂

∂u3(280)

Therefore:

∇ (ΦA) =

1h1

∂∂u1

1h2

∂∂u2

1h3

∂∂u3

ΦA1ΦA2ΦA3

=

Φ

h1

∂A1

∂u1+

A1

h1

∂Φ

∂u1+

Φ

h2

∂A2

∂u2+

A2

h2

∂Φ

∂u2+

Φ

h3

∂A3

∂u3+

A3

h3

∂Φ

∂u3

=(∇Φ) A + Φ(∇ A)

(281)

11 References

[1] Michael Artin , “Algebra”, Prentice Hall, 1991[2] Richard Courant and David Hilbert , “Methods of Mathematical Physics Volume 1”, Wiley,1989[3] I M Gelfand and S V Fomin , “Calculus of Variations”, translated and edited by Richard A Silverman,Dover,1991[4] Oliver D Kellogg , “Foundations of Potential Theory”, Dover,1953[5] Fritz John, “Partial Differential Equations - Third Edition”, Springer,1978[6] Erwin Kreyszig, “Differential Geometry”, Dover,1991[7] P.S.Laplace, “A Treatise of Celestial Mechanics”, Translated by Rev Henry H Harte, printed by RichardMilliken, Dublin, 1822 https://archive.org/details/treatiseofcelest12lapl[8] Peter J Olver , “Introduction to Partial Differential Equations”, Springer, 2014[9] H M Schey , “Div, Grad, Curl and all that - an informal text on vector calculus”, First Edition,W WNorton & Company, 1973

12 History

Created 22/07/2018

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